In fig. 6. 36, QR/QS =QT/PR …
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Sia ? 6 years, 6 months ago
Given: In figure, {tex}\frac{{QR}}{{QS}} = \frac{{QT}}{{PR}}{/tex}and {tex}\angle{/tex} 1 = {tex}\angle{/tex} 2
To prove: {tex}\triangle PQS \sim \triangle TQR{/tex}
Proof: In {tex}\triangle {/tex} PQR {tex}\because {/tex} {tex}\angle{/tex} 1= {tex}\angle{/tex} 2
{tex}\therefore {/tex} PR = QP (1).......[ {tex}\because {/tex} sides opposite to equal angle of a triangle are equal]
Now, {tex}\frac{{QR}}{{QS}} = \frac{{QT}}{{PR}}{/tex} ......given
{tex} \Rightarrow \frac{{QR}}{{QS}} = \frac{{QT}}{{QP}}{/tex} (2).......Using(1)
Again in {tex}\triangle PQS{/tex} and {tex}\triangle TQR{/tex}
{tex}\because \frac{{QR}}{{QS}} = \frac{{QT}}{{QP}}...........From(2){/tex}
{tex}\therefore \frac{{QS}}{{QR}} = \frac{{QP}}{{QT}}{/tex} and {tex}\angle SQP = \angle RQT{/tex}
{tex}\therefore \triangle PQS \sim \triangle TQR{/tex}...........SAS similarity criterion
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